In the morning, as the sun climbs in the sky, my shadow gets shorter and shorter. In the afternoon, as sun gets lower and lower in the sky, my shadow grows again. At local noon, when the sun is at its highest, my shadow is as short as it’s going to be that day. My noon-time shadow is my shortest shadow.

How high the sun gets at local noon depends on the time of year. In the summer, it gets much higher than in the winter. On the day of the winter solstice, the noon-time sun is lower than the noon-time sun on any other day of the year. Therefore, of all of my shortest (noon-time) shadows, the one on the winter solstice will be the longest. It’s my longest shortest shadow.

Today is the winter solstice, and I’d hoped to photograph my longest shortest shadow on the beach at Portobello in Edinburgh. But implicit in all of the above is that the sun will be visible in the sky at the critical moment in order to cast the shadow. In Scotland in December this can be a tricky condition to meet, and it wasn’t quite met today. My shadow was barely visible. Since this idea came to me, I’ve only had one chance to photograph my longest shortest shadow, back in 2010.

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The hollow face illusion is a wonderful visual effect in which a hollow mask of a face appears to be convex, like the face itself. Making a hollow mould of your face (for example using plaster) is difficult and potentially dangerous. However, last weekend my attention was drawn to an easier and safer way.

I was walking down from Coire an Lochain in the Scottish Highlands with a group from the Red Rope club, when I saw my friend Maia standing on the path ahead, chuckling. She’d been making face imprints in a steep snowdrift, and they showed the hollow face illusion beautifully.

The procedure needs no explanation (see right). The snow needs to be fresh and soft; you’d be surprised how hard it is to push your face into what feels to your hand like very soft snow. The tip of my nose is noticeably flattened in the picture above.

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Cut Adrift is an installation by Edinburgh artist Mark Haddon. Over the past 20 years or so, Mark has collected handwritten notes that he has found lying on the ground. The notes are a varied mix: letters, recipes, instructions, apologies… Mark recorded people reading these notes out loud. As part of Sanctuary, a 24-hour art event in southern Scotland, he and I arranged that the passing of people along a path would trigger the playing of these recordings from a sound system hidden in the undergrowth.

I helped Mark with the technical side of the project. We used a Teensy microcontroller board with an audio adaptor board audio shield to store, sequence, and play the audio clips. The signal went to a 12 V audio amplifier and thence to a pair of loudspeakers. We used a PIR (passive infra-red) motion detector to detect people walking by. The output from the detector was connected to one of the input pins of the Teensy. The whole thing was powered using a 12 V lead-acid battery. The battery, Teensy, and amplifier all went in a plastic storage crate to protect them from the weather.

The passage of a person triggered one audio clip. The choice of which clip to play was random, but subject to a rule that made a clip more and more likely to be chosen the longer it was since it was last played. My program held the clips in a queue. When a clip was needed, there was a 50% probability that the first clip in the queue would be chosen, a 25% probability that the second clip would be chosen, and so on as far as the 5th clip. Clips lower in the queue than this would never be chosen. Once the chosen clip had been played, it was put at the bottom of the queue and all the other clips moved up one place.

I’m grateful to Jen Sykes of Glasgow School of Art for pointing me in the direction of the Teensy and its audio board.

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His printer provided a sample run so that we knew what the areal density of the ink film would be, and I did the sums to work out how big the squares would need to be and how many pages we’d need. It turned out that we needed squares of side 19.28 cm, and (coincidentally) 1928 pages.

The responsibility weighed heavily on me, and I was a little nervous as the order went off to the printer. So I was relieved and delighted when Edwin told me that when he’d walked into the printers to pick the job up, the printer said ‘bloody hell, it took a whole kilo tin of ink to print your job’.

Edwin is Lecturer, Typography Technician and Designer in Residence at Glasgow School of Art.

]]>In the previous post, we discovered that the kinetic energy of a drifting continent is of the same general magnitude as that of a moving bicycle and its rider – 1500 joules would be a typical figure.

I went on to calculate that, whereas it takes me only about 10 seconds to get my bike up to full speed, it would take me hundreds of years to get the continent up to its tiny full speed were I to put my shoulder against it and push (assuming that it was perfectly free to move). How can this be, when the amount of energy that I’m giving each of these objects is the same?

The problem is that when I push the continent, I am, effectively, in the wrong gear.

On a bike with gears, you’ve got a range of choices about how you power it: you can ride in a high gear, pedalling slowly but pushing hard on the pedals, or ride in a low gear, pedalling more quickly but pushing less hard on the pedals. There’s a simple tradeoff: if you want to pedal half as fast, you’ve got to push twice as hard for the same effect.

But there’s a limit to how hard you can push on the pedals, which means that if you move up too far up through the gears, there comes a point where you can no longer make up for the decreased pedalling rate by pushing harder on the pedals, and the power that you can supply to the bicycle falls.

Anyone who’s tried to accelerate a bicycle when they are in too high a gear will have experienced this problem, and it’s what I experience when I try to push the continent directly. Because the top speed of the continent is extremely low (about the speed of a growing fingernail), I’m necessarily pushing it very slowly as I accelerate it. This means that to give it energy at the rate that I want to (1500 joules in 10 seconds, like the bike) I would have to push it impossibly hard – the force needed is about the same as the weight of a 300-metre cube of solid rock.

Is there a way that we can put me into a lower gear, so that I can push with a force that suits me, over a longer distance, and still apply the very high force over a short distance to the continent?

Yes. Just as we’ve all used a screwdriver as a lever to get the lid of a tin of paint off, so I could use a lever to move the continent. Similarly to the bike gears, the lever allows me to exchange pushing hard over a small distance with pushing less hard over a longer distance. To do the job, the lever would need to be long enough to allow me to push, with all my might, through a distance of about 2.5 metres, with the short arm of the lever pushing the continent. We’d need an imaginary immoveable place for me to stand, and we could use the edge of the neighbouring continent as the pivot (just as we use the rim of a paint tin as the pivot). The catch is the length of the lever: if the short arm was 1 metre long, the long arm would be about 1.5 million kilometres long.

Simon Gage of Edinburgh International Science Festival suggested a more compact arrangement: a bicycle with an extremely low gear ratio, with the front wheel immobilised on the neighbouring continent (assumed immoveable), and the back wheel resting on the continent we’re trying to accelerate. A transmission giving 17 successive 4:1 speed reductions would do the job nicely. Ten seconds of hard pedalling would get the continent up to full speed. To me on the saddle, it shouldn’t feel any different to accelerating my bike away from the lights.

*A wee caveat*. This is a thought experiment, and we’ve swept some fairly significant engineering issues under the carpet. The rearmost parts of the power train would be moving at speeds that are literally geological, so in reality it would take me years of pedalling to take all of the slack and stretch out of the system. These parts would also be transmitting mountainous forces, and so they’d need to be supernaturally strong. There will be frictional losses. And then there’s the issue of transmitting a gigantic force to the continent through the contact of a bike tyre on the ground.

What force is required to accelerate the Eurasian plate to top speed in 10 seconds?

The top speed of the plate is 3.2 × 10^{-10} ms^{-1}. If I accelerate it uniformly, its average speed will be half of this, and so in the 10 seconds over which I hope to accelerate it, it will travel 1.6 × 10^{-9} m.

Now *W = fd*

where *W* is the work that I do on the plate (ie the kinetic energy that I give it), *f* is the force that I apply to it, and *d* is the distance through which I push the plate. Rearranging gives us

*f = W/d*

We know *W* from the previous post (it’s 1500 joules) and we’ve just calculated *d*. Thus *f* works out at about 9.4 × 10^{11} newtons.

For comparison, a 300-metre cube of rock of density 2700 kg m^{-3} will have a weight of (300 m)^{3} × 2700 kg m^{-3} × 9.81 m s^{-2} = 7 × 10^{11} newtons roughly.

When a lever is used to amplify a force, the ratio of the lengths of the arms of the lever needs to be the same as the ratio of the two forces. Suppose that I can push with a force equal to my own body weight, about 600 newtons. If I’m to use a lever to amplify my push of 600 N to a force of 9.4 × 10^{11} N, the ratio of the lengths of the arms needs to be (9.4 × 10^{11})/600, or roughly 1.5 × 10^{9}. So if the short arm of the lever is 1 metre long, the long arm needs to be about 1.5 × 10^{9} metres long, which is 1.5 million kilometres. For comparison, the Moon is about 400,000 kilometres away.

To do 1500 joules of work with a force of 600 N, I’d need to push over a distance of 2.5 metres (because 600 × 2.5 = 1500).

I estimated that it takes me 15 pedal revolutions to get my bike up to full speed. Knowing the length of the pedal cranks, I know the total distance that I have pushed the pedals through, and I know how much work I have done on the bicycle – 1500 joules. (I’m ignoring energy losses here, because they are small at low speeds on a bike and the calculation is highly approximate anyway). Using *work done = force × distance*, this gives an average force on the pedals of about 94 newtons.

The 17 stages of 4:1 reduction mean that the back wheel is rotating 417 = 1.7 × 10^{10} times slower than I’m pedalling. The pedalling force is amplified in the same ratio, to give a force on the teeth of the rearmost gear of 1.6 × 10^{12} newtons. We now have to allow for the fact that the radius of the rear wheel is about twice the length of the pedal crank. This roughly halves the force available at the rim of the rear wheel, giving a force of about 8 × 10^{11} newtons, which is close to what we need.

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According the the theory of plate tectonics, the outer layer of the Earth is divided into a number of separate plates, which very slowly drift around, opening and closing oceans, causing earthquakes, and thrusting up mountain ranges.

A moving body has energy by virtue of its motion: kinetic energy. Kinetic energy is proportional to a body’s mass and to the square of its speed.

Now tectonic plates move extremely slowly: the usual comparison is with a growing fingernail. But they are also extremely heavy: tens of millions of square kilometres in area, over 100 km thick, and made of rock. I wondered how the minute speed and colossal mass play out against each other: what’s the kinetic energy of a drifting tectonic plate?

There are so many variables, that vary such a lot, that this calculation is going to be extremely approximate. But the answer is delightfully small: the kinetic energy of the tectonic plate on which I live, as observed from one of the plates next door, is about the same as the kinetic energy of me and my bike when I’m going at a reasonable pace: about 1500 joules.

This is a fun calculation to do, but we shouldn’t get carried away thinking about the kinetic energy of tectonic plates. Plates are driven by huge forces, and their motion is resisted by equally large forces. The mechanical work done by and against these forces will dominate a plate’s energy budget in comparison to its kinetic energy.

But the calculation does provoke an interesting thought about forces and motion. I can get my bike up to full speed in, say, 10 seconds. If the Eurasian plate were as free to move as my bike, and I were to put my shoulder against it and shove as hard as I could, it would take me about 500 years to get it up to its (very tiny) full speed.

In both cases, I’m giving the moving object roughly 1500 joules of kinetic energy. How come I can give that energy to my bike in a few seconds, but to give it to the plate would take me centuries?

I’ll return to that thought in a later post.

Depending on how you count them, there are 6-7 major tectonic plates, 10 minor plates, and many more microplates. The plates vary hugely in size, from the giant Pacific Plate with an area of 100 million km^{2}, to the dinky New Hebridean plate, which is a hundred times smaller. The microplates are smaller still. Plates also vary a lot in speed: 10-40 mm is typical.

I’m going to be parochial, and choose the Eurasian plate for this calculation.

Let’s call the area of the plate *a* and its mean thickness *t*. Its volume is then given by *at*, and if its mean density is *ρ*, then its mass *m* is *ρat.*

A body of mass *m * moving at a speed *v *has kinetic energy ½*mv*^{2}. So our plate will have kinetic energy ½*ρatv*^{2}.

The area of the Eurasian plate is 67,800,000 km^{2} or 6.78 × 10^{13} m^{2}, and its speed relative to the African plate is (the only speed I have) is given as 7-14 mm per year. We’ll use 10 mm per year, which is 3.2 × 10^{-10} ms^{-1}. The thickness of tectonic plates in general varies roughly in the range 100-200 km depending upon whether we are talking about oceanic or continental lithosphere; let’s call it 150 km or 1.5× 10^{5} m. The density of lithospheric material varies in the range 2700-2900 kg m^{-3}; we’ll use 2800 kg m^{-3}.

Putting all of these numbers into our formula for kinetic energy, we get a value of 1500 joules (to 2 significant figures, which the precision of the input data certainly doesn’t warrant).

Now for me and my bike. I weigh about 57 kg, my bike is probably about 10 kg. Suppose I’m riding at 15 mph, which is 6.7 ms^{-1}. My kinetic energy is almost exactly…

…1500 joules!

The closeness of these two values is unmitigated luck*, and we shouldn’t be seduced by the coincidence. Just varying the speed of the plate in the range 7-14 mm would cause a 4-fold change in kinetic energy, and there’s the variability in plate thickness and rock density to take into account as well. The choice of bike speed was arbitrary, I guessed the mass of the bike, and I’ve since realised that I didn’t account for the fact that the wheels of my bike rotate as well as translate.

However, what we can say is that the kinetic energy of a drifting continent is definitely on a human scale, which leads to a new question:

Suppose the Eurasian plate were as free to move as my bicycle, and that I put my shoulder against it and shoved, how long would it take me to get it up to speed?

From the figures above, the mass of the plate is 2.85 × 10^{22} kg. If I can push with a force equal to my own weight (about 560 newtons) then by Newton’s 2nd Law I can give it an acceleration of about 1.96 × 10^{-20} ms^{-2}. Rearranging the equation of motion *v = at*, where *v* is the final speed, *a* is the acceleration, and *t* is the time, then *t = v/a. *Inserting the values for *v *and *a*, we get *t = *1.6 × 10^{10} seconds, or about 500 years.

* I didn’t tweak my assumptions: what you see above really is the very first version of the calculation!

]]>On 16th April Sarah McLeary and I had a busy day at the 2017 Mini Maker Faire in Edinburgh (part of the science festival). We showed the paraboloidal castings that we’ve been working on together, and I showed developments in my irregular polyhedra since I showed them at the 2015 Mini Maker Faire.

As well as plaster casts, we’ve tried slipcasting porcelain using our paraboloidal moulds, squirting the slip on as the mould spins to try and get a lacy structure. You can see one of these near the front left of the table. I’ve also been having a go at making irregular puckered plane-based tilings (standing up at the back of the table). I will write more about both of these projects before too long.

]]>When you stir a cup of tea, the surface of the rotating liquid develops a dip in the middle. The faster you stir, the deeper the dip. But the liquid surface is quite uneven; to get a smooth surface, throw the spoon away and spin the whole cup continuously. Once the liquid inside has caught up with the cup, and everything is turning at the same speed, the liquid surface forms a beautiful smooth curve known as a paraboloid of revolution.

Sarah McLeary and I are applying this idea to make thin paraboloidal plaster shells. We spin a bucket on a potter’s wheel (Sarah is a potter), and pour plaster into it. The plaster rapidly flows until its surface forms a deep paraboloidal curve, and then sets. We now use this cast, still spinning, as a mould, and cast a thin layer of plaster inside it, to make a paraboloidal shell.

That’s our first attempt above. It’s about 20 cm across and 3 mm thick. It might look like a part of a sphere, but in a profile view it’s easy to see that the curvature is tightest at the base and gradually decreases up the sides, as you’d expect for a paraboloid.

I love the fact that we didn’t decide what shape this shell was going to be: physics did.

There’s lots of experimentation ahead. When we’ve got the hang of it, I’ll explain our methods in more detail. But as this picture of our third attempt shows, we haven’t quite cracked it yet.

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This image is my version of Edward Adelson’s checkershadow illusion (with a little inspiration from Magritte). It’s a photograph of a real, physical scene.

Take a look at the central square of the checkerboard, and the square indicated by the arrow. Which is lighter? Quite clearly, it’s the central square, isn’t it?

Remarkably, the central square actually emits *less* light than the square indicated by the arrow! You could use a light meter to check this claim, but it’s easier to verify it directly by using a piece of card with two holes cut in it to mask off the rest of the image.

Some people will tell you that this image shows you how easy it is to fool your brain. But it does the exact opposite: it shows you what a marvellous piece of equipment your brain is.

Think about the checkerboard itself, and the materials it’s made of. The arrowed square is coated with dark grey paint, and the central square is coated with light grey paint—and that’s exactly what you perceive.

The shadow cast by the pipe means that the light-grey central square is more dimly lit than the dark-grey arrowed square, so much so that it actually reflects less light into your eye than the arrowed square. But your brain cleverly manages to determine the actual lightnesses of the physical surfaces, despite the uneven lighting. Isn’t that a *good* thing for your brain to do?

If you still don’t believe me, try this thought experiment. Imagine that you live in a forest where there are two kinds of fruit. One is light grey and poisonous, and the other is dark grey and nutritious. Two of these fruits hang next to each other, but in the dappled forest light the (light grey) poisonous fruit is in shadow, and the (dark grey) nutritious fruit is in bright light. Suppose that the depth of the shadow is such that the light-grey poisonous fruit actually reflects slightly *less* light into your eye than the dark-grey nutritious fruit, just as with the two squares in the picture above. Would you *really* want your vision to tell you that the poisonous fruit was the dark one and therefore the one to pick? Or would you want it to discount the irrelevant effect of the shadow and tell you which fruit was *actually *dark and which was *actually* light (and would kill you)? I know what I’d want.

I think that it is wrong to call this effect an illusion (and so does Adelson). There is nothing illusory about what you see. You perceive the useful truth about the scene in front of you.

]]>That’s what we’re often told, and boiling only as much water as you need for your cup of tea can only be a good thing. But how much impact does it *really* have on your overall energy consumption?

Consider this: if you’re driving along in a car at 50 mph, you’re using enough energy to make a cup of tea *every few seconds*.

1, 2, 3, tea, 1, 2, 3, tea, 1, 2, 3, tea…

As most people don’t think twice about driving their cars for an extra few minutes, never mind an extra few seconds, it’s clear that the energy savings made by being frugal with your kettle are very small compared to your overall energy use.

Does this mean that we shouldn’t be careful with our kettles after all? No. It all matters. But having boiled exactly one cup’s worth of water for your refreshing cuppa, you shouldn’t put your feet up and think that your energy consumption issues are sorted. You’ve *much* bigger fish to fry.

Imagine that you’re travelling in a car that’s doing 50 miles per hour, with a rate of fuel consumption of 50 miles to the gallon (the units are customary in a UK motoring context). This is a realistic situation; if anything it’s optimistic in terms of energy consumption.

In this scenario, your car will burn a gallon of petrol, that’s 4.5 litres approximately, every hour.

With 3600 seconds in an hour, that’s 4.5L/3600s = 0.00125 litres per second. The density of petrol is about 0.8 kg per litre, so the rate of petrol consumption comes to 0.00125 L s^{-1} × 0.8 kg L^{-1} = 0.001 kilograms per second.

The energy density of hydrocarbon fuels is about 46 megajoules per kilogram, so the rate of use of energy by your car is 0.001 kg s^{-1} × 46×10^{6} J kg^{-1} = 46000 joules per second (watts).

Now let’s think about the tea. The energy *E *required to produce a temperature change of Δ*T *in a mass *m* of a substance whose specific heat capacity is *C* is

*E* = *m C* Δ*T *

A typical mug of tea contains about 250 ml, or 0.25 kg, of water. For water, *C *= 4200 J kg^{-1} K^{-1}. The temperature change is about 90 K, if the water comes out of the tap at about 10°C and is raised to boiling point. So the amount of energy required to heat the water for a cup of tea is 0.25 kg × 4200 J kg^{-1} K^{-1}× 90 K = 94500 joules.

As the car is using 46000 joules per second, it follows that the car is using enough energy to heat the water for a cup of tea every two seconds.

Now to be fair to the car, we need to recognise that in real life there is more energy used to heat the water for the tea than actually goes into the water. There are heat losses in energy generation and transmission, and from the kettle itself.

Things get very complicated here. A thermal electricity generating plant (coal, oil, gas, or nuclear) converts energy in the fuel to electrical energy with about 35-45% efficiency. But in the UK, about 20% of our electricity is generated from renewables, where the concept of efficiency is harder to pin down. For example, I could use some oil to heat my water directly instead of doing it indirectly by generating electricity, but I couldn’t do the same with the wind. I’m going to assume that 80% of the electricity that I used was generated in thermal plants at 40% efficiency, and 20% of it was generated at 100% efficiency, giving an overall efficiency of 52%. But be aware that this is a *very* rough figure, and that there is no right answer.

According to this document, about 93% of the energy in generated electricity makes it unscathed through the transmission and distribution systems to the end user.

I did an experiment and estimated the efficiency of my kettle to be about 88%. I give details later.

This gives an overall efficiency of boiling water as 52% × 93% × 88% = 43%. So instead of using 94500 joules to boil water for a cup of tea, we’re actually using 94500J/0.43 = 220000 joules, which means that the car’s rate of energy use is equivalent to a cup of tea roughly every 5 seconds. Because of all the uncertainties involved (not least the size of a cup of tea), we should treat this figure as very approximate.

My kettle took 180 s to raise 1 litre of water from 17°C to boiling.

The kettle is rated at 1850-2200 W for supply voltages in the range 220-240 V. When the kettle was running, I measured the supply voltage to be 242 V, so I assumed that the kettle was operating at the top of its power range, ie 2200 W.

The heat supplied by the kettle element was therefore 2200 W × 180 s = 396000 J.

Using the same equation as earlier, the energy required to raise the temperature of 1 kg of water by 83 K is 1 kg × 4200 J kg^{-1} K^{-1} × 83 K = 348600 J.

The kettle was therefore heating the water with efficiency 348600/396000 = 0.88 or 88%.

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